Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $z \neq 0$. $q = \dfrac{z^2 - 19z + 90}{-2z^3 + 26z^2 - 60z} \div \dfrac{2z + 20}{5z^2 - 15z} $
Dividing by an expression is the same as multiplying by its inverse. $q = \dfrac{z^2 - 19z + 90}{-2z^3 + 26z^2 - 60z} \times \dfrac{5z^2 - 15z}{2z + 20} $ First factor out any common factors. $q = \dfrac{z^2 - 19z + 90}{-2z(z^2 - 13z + 30)} \times \dfrac{5z(z - 3)}{2(z + 10)} $ Then factor the quadratic expressions. $q = \dfrac {(z - 10)(z - 9)} {-2z(z - 10)(z - 3)} \times \dfrac {5z(z - 3)} {2(z + 10)} $ Then multiply the two numerators and multiply the two denominators. $q = \dfrac { (z - 10)(z - 9) \times 5z(z - 3)} { -2z(z - 10)(z - 3) \times 2(z + 10)} $ $q = \dfrac {5z(z - 10)(z - 9)(z - 3)} {-4z(z - 10)(z - 3)(z + 10)} $ Notice that $(z - 10)$ and $(z - 3)$ appear in both the numerator and denominator so we can cancel them. $q = \dfrac {5z\cancel{(z - 10)}(z - 9)(z - 3)} {-4z\cancel{(z - 10)}(z - 3)(z + 10)} $ We are dividing by $z - 10$ , so $z - 10 \neq 0$ Therefore, $z \neq 10$ $q = \dfrac {5z\cancel{(z - 10)}(z - 9)\cancel{(z - 3)}} {-4z\cancel{(z - 10)}\cancel{(z - 3)}(z + 10)} $ We are dividing by $z - 3$ , so $z - 3 \neq 0$ Therefore, $z \neq 3$ $q = \dfrac {5z(z - 9)} {-4z(z + 10)} $ $ q = \dfrac{-5(z - 9)}{4(z + 10)}; z \neq 10; z \neq 3 $